Approaching Asymptotes of Hyperbolas

Date: 04/24/2011 at 08:10:04
From: Donald
Subject: Derivations of standard hyperbolas asymptotes
How do you derive the equations for the asymptotes of the standard
hyperbolas?
y = +/-(b/a)x
y = +/-(a/b)x
Solving for y, I got it down to:
y = +/-(b/a) sqrt(x^2 - a^2)
Then, letting x go to infinity, the a^2 is rendered insignificant, so
y = +/-(b/a) sqrt((x^2))
This gives
y = +/-(b/a)x
But, how does this prove that y = +/- (b/a)x is an asymptote(s)? I need
clarity here.
I have been to site after site, and looked in books, and I still can't
find an explanation. They all just state it and how to use it, but offer
no proof. Could you help me with this?

Date: 04/24/2011 at 22:46:44
From: Doctor Peterson
Subject: Re: Derivations of standard hyperbolas asymptotes
Hi, Donald.
What you've done is a good informal demonstration of the idea; we can make
it a little more convincing by saying it this way:
y = +/-b/a sqrt(x^2 - a^2)
= +/-b/a sqrt(x^2(1 - a^2/x^2))
When x is much larger than a, a^2/x^2 is much less than 1, so this will be
very close to
y = +/-b/a sqrt(x^2) = +/-(b/a)x
For a real proof, we have to start with the definition of "asymptote";
without that, no proof is possible, since we wouldn't know what we were
trying to prove!
An asymptote is a line that is approached more and more nearly by the
curve as x increases. That is, if we have a curve y = f(x) and a line
y = mx + b, the latter is an asymptote of the former if
lim[x->oo](f(x) - (mx + b)) = 0
To show that y = bx/a is an asymptote of y = b/a sqrt(x^2 - a^2), we want
to show that
lim[x->oo](b/a sqrt(x^2 - a^2) - bx/a) = 0
Is it?
lim[x->oo](b/a sqrt(x^2 - a^2) - bx/a)
= lim[x->oo](b/a (sqrt(x^2 - a^2) - x))
= b/a lim[x->oo](sqrt(x^2 - a^2) - x)
(sqrt(x^2 - a^2) - x)(sqrt(x^2 - a^2) + x)
= b/a lim[x->oo]--------------------------------------------
(sqrt(x^2 - a^2) + x)
(x^2 - a^2)^2 - x^2
= b/a lim[x->oo]---------------------
sqrt(x^2 - a^2) + x
x((1 - a^2/x^2)^2 - 1)
= b/a lim[x->oo]--------------------------
x(sqrt(1 - a^2/x^2) + 1)
(1 - a^2/x^2)^2 - 1
= b/a lim[x->oo]-----------------------
sqrt(1 - a^2/x^2) + 1
(1)^2 - 1
= b/a lim[x->oo]-------------
sqrt(1) + 1
0
= ---
2
= 0
So the curve does in fact approach the line.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

Date: 04/26/2011 at 06:34:47
From: Donald
Subject: Thank you (Derivations of standard hyperbolas asymptotes)
I understand, and thank-you so much for the precision and the clarity.
Blessings to you and yours,
D.